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Formulas/maths/M2 Modulus Argument/Triangle Inequality

Triangle Inequality

The modulus of a sum never exceeds the sum of moduli. Geometrically: one side of a triangle is at most the sum of the other two. Equality holds iff z₁ and z₂ have the same argument — i.e., they point in the same direction from the origin.
Derivation

The triangle inequality says the length of one side of a triangle cannot exceed the sum of the other two. In the complex plane, this becomes a statement about moduli.

Expanding ∣z1+z2∣2|z_1 + z_2|^2

Start by writing ∣z1+z2∣2|z_1 + z_2|^2 as a product with its conjugate:

∣z1+z2∣2=(z1+z2)(z1+z2)‾=(z1+z2)(zˉ1+zˉ2)|z_1 + z_2|^2 = (z_1 + z_2)\overline{(z_1 + z_2)} = (z_1 + z_2)(\bar{z}_1 + \bar{z}_2)

Expanding:

=z1zˉ1+z1zˉ2+z2zˉ1+z2zˉ2= z_1\bar{z}_1 + z_1\bar{z}_2 + z_2\bar{z}_1 + z_2\bar{z}_2 =∣z1∣2+z1zˉ2+z1zˉ2‾+∣z2∣2= |z_1|^2 + z_1\bar{z}_2 + \overline{z_1\bar{z}_2} + |z_2|^2

The middle two terms are a complex number and its conjugate, so their sum is twice the real part:

∣z1+z2∣2=∣z1∣2+2Re⁡(z1zˉ2)+∣z2∣2|z_1 + z_2|^2 = |z_1|^2 + 2\operatorname{Re}(z_1\bar{z}_2) + |z_2|^2

Applying the Key Bound

For any complex number ww:

Re⁡(w)≤∣w∣\operatorname{Re}(w) \leq |w|

This is simply because Re⁡(w)\operatorname{Re}(w) is one side of a right triangle with hypotenuse ∣w∣|w|.

Applying this with w=z1zˉ2w = z_1\bar{z}_2:

Re⁡(z1zˉ2)≤∣z1zˉ2∣=∣z1∣∣zˉ2∣=∣z1∣∣z2∣\operatorname{Re}(z_1\bar{z}_2) \leq |z_1\bar{z}_2| = |z_1||\bar{z}_2| = |z_1||z_2|

Substituting back:

∣z1+z2∣2≤∣z1∣2+2∣z1∣∣z2∣+∣z2∣2=(∣z1∣+∣z2∣)2|z_1 + z_2|^2 \leq |z_1|^2 + 2|z_1||z_2| + |z_2|^2 = (|z_1| + |z_2|)^2

Taking square roots of both sides (both sides are non-negative):

∣z1+z2∣≤∣z1∣+∣z2∣\boxed{|z_1 + z_2| \leq |z_1| + |z_2|}

Equality Condition

Equality holds iff Re⁡(z1zˉ2)=∣z1zˉ2∣\operatorname{Re}(z_1\bar{z}_2) = |z_1\bar{z}_2|, which happens iff z1zˉ2z_1\bar{z}_2 is real and non-negative.

z1zˉ2z_1\bar{z}_2 is real and non-negative iff arg⁡(z1zˉ2)=0\arg(z_1\bar{z}_2) = 0 iff arg⁡(z1)−arg⁡(z2)=0\arg(z_1) - \arg(z_2) = 0 iff arg⁡(z1)=arg⁡(z2)\arg(z_1) = \arg(z_2).

Equality holds iff z1z_1 and z2z_2 point in the same direction from the origin — i.e., one is a positive real multiple of the other.

Geometrically: equality holds when the three points 00, z1z_1, z1+z2z_1 + z_2 are collinear with z1z_1 and z2z_2 on the same side — the parallelogram degenerates to a line segment.

The Reverse Triangle Inequality

From the same expansion, replacing z2z_2 with −z2-z_2:

∣z1−z2∣≥∣∣z1∣−∣z2∣∣|z_1 - z_2| \geq \bigl||z_1| - |z_2|\bigr|

This follows because ∣z1∣=∣(z1−z2)+z2∣≤∣z1−z2∣+∣z2∣|z_1| = |(z_1 - z_2) + z_2| \leq |z_1 - z_2| + |z_2|, giving ∣z1−z2∣≥∣z1∣−∣z2∣|z_1 - z_2| \geq |z_1| - |z_2|. Swapping z1z_1 and z2z_2 gives the other side of the absolute value.